ZOJ Problem Set - 2109 FatMouse' Trade

  FatMouse' Trade

---

Time Limit: 2 Seconds      Memory Limit: 65536 KB

---

FatMouse prepared M pounds of cat food, ready to trade with the cats guarding the warehouse containing his favorite food, JavaBean.
The warehouse has N rooms. The i-th room contains J[i] pounds of JavaBeans and requires F[i] pounds of cat food. FatMouse does not have to trade for all the JavaBeans in the room, instead, he may get J[i]* a% pounds of JavaBeans if he pays F[i]* a% pounds of cat food. Here a is a real number. Now he is assigning this homework to you: tell him the maximum amount of JavaBeans he can obtain.

Input

The input consists of multiple test cases. Each test case begins with a line containing two non-negative integers M and N. Then N lines follow, each contains two non-negative integers J[i] and F[i] respectively. The last test case is followed by two -1's. All integers are not greater than 1000.

Output

For each test case, print in a single line a real number accurate up to 3 decimal places, which is the maximum amount of JavaBeans that FatMouse can obtain.

Sample Input

5 3
7 2
4 3
5 2
20 3
25 18
24 15
15 10
-1 -1

Sample Output

13.333
31.500

---

Author: CHEN, Yue

Source: Zhejiang Provincial Programming Contest 2004 

分析：

题意：

给定若干数据，每个数据包含两个值，一个是a，一个是b，性价比是a/b。要求每次取性价比高的，且b的和不能超过给定的m。输出a的和的最大值。

简单题。可分割的背包问题，贪心策略，不需用到dp就能解决。

重点练习结构体排序，终于搞懂了结构体排序的原理。么么哒。

ac代码：

#include <iostream>
#include<cstdio>
#include<algorithm>
using namespace std;
const int maxm=1000;
const int maxn=1000;
struct Mouse
{
    double a,b;
    bool operator <(const Mouse &m2) const//运算符重载，注意没有>运算符重载，要表示大于关系，只需改变重载函数下面的符号
    {
        //return m2.a/m2.b<a/b;//由大到小排序，用<;由小到大排序，用>
        return a/b>m2.a/m2.b;//由大到小排序，用>;由小到大排序，用<。这种表示法更直观
    }
}mouse[maxm];
int main()
{
    int n,i,j;
    double m;
    while(scanf("%lf%d",&m,&n)&&m!=-1&&n!=-1)
    {
        for(i=0;i<n;i++)
        scanf("%lf%lf",&mouse[i].a,&mouse[i].b);
        sort(mouse,mouse+n);
        //for(i=0;i<n;i++)
        //cout<<mouse[i].a<<"  "<<mouse[i].b<<"  "<<mouse[i].a/mouse[i].b<<"  "<<endl;//测试
        double s=0;
        for(i=0;i<n&&m>0;i++)
        {
            if(m>=mouse[i].b)
            {
                s+=mouse[i].a;
                m-=mouse[i].b;
            }
            else if(m<mouse[i].b)
            {
                s+=mouse[i].a*m/mouse[i].b;
                m=0;
            }
        }
        printf("%.3lf\n",s);
    }
    return 0;
}